Thevenin equivalent of a circuit

In summary, the problem asks for the voltage Vo across an isolated voltage source and current source in the left part of the circuit. You cannot find Vo using Thevenin's theorem because you do not know the voltage drop across the current source. You can only find Vo by first calculating the Thevenin equivalent of the voltage source and capacitor, then marching across the circuit to find the result.
  • #1
aid
16
0

Homework Statement



I am attempting to solve problem number 4 from the following picture:

http://img641.imageshack.us/img641/7950/imageisx.jpg

In the picture you can see the suggested line of "cut" for the first application of Thevenin's theorem.


The Attempt at a Solution



I have tried to find the Thevenin equivalent voltage (for the part on the left from the sketched line). I get:
[tex]V_{oc} = 12 \angle 0 - 2 * 6 \angle 0 = 0[/tex]
which would render the whole exercise surprisingly trivial.

What the heck is wrong with the above equation? I would be grateful for any help on this one.
 
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  • #2
KVL in left part is not possible, if that is what you are doing ! KVL is not possible because of the presence of current source across which you don't know the voltage drop !
 
  • #3
Hi Aid. Is that circle with + and | marked on it supposed to represent an independent voltage source? If so, it fixes the voltage across the network of passive components to its right.
 
  • #4
lazyaditya said:
KVL in left part is not possible, if that is what you are doing ! KVL is not possible because of the presence of current source across which you don't know the voltage drop !

What method is possible to apply in this case, then? Is it really so that the Thevenin equivalent voltage of the marked part is equal to:

[tex]V_{oc} = 12 \angle0 V[/tex]?
 
  • #5
aid said:
What method is possible to apply in this case, then? Is it really so that the Thevenin equivalent voltage of the marked part is equal to:

[tex]V_{oc} = 12 \angle0 V[/tex]?

A fixed voltage source effectively isolates branches parallel to it. In this case the six amp current supply and its 2 Ω series resistor cannot affect the voltage across that voltage source in any way. So, if you are looking to find the voltage Vo which is to the right of the voltage source, you can discard anything to the left of that isolating supply from consideration. That is, ignore entirely the 6 Amp supply and its 2 Ω series resistance; the 12V voltage source will be the starting point for your march across the circuit towards Vo.

The question requires you to employ Thevenin's theorem to find the result, so I'd suggest applying it several times, accumulating chunks of the circuit into successive Thevenin models (source voltage and series impedance) as you go. Where you've drawn your first "cut" in pencil is a good staring point -- the Thevenin equivalent of the voltage source and capacitor alone is very straightforward.
 

FAQ: Thevenin equivalent of a circuit

What is the Thevenin equivalent of a circuit?

The Thevenin equivalent of a circuit is a simplified representation of a complex circuit that can be used to analyze and predict the behavior of the original circuit. It consists of a voltage source in series with a resistor connected across two terminals, and is equivalent to the original circuit in terms of voltage and current at those terminals.

How is the Thevenin equivalent calculated?

The Thevenin equivalent is calculated by finding the open-circuit voltage and the equivalent resistance of the original circuit. The open-circuit voltage is the voltage across the two terminals when they are disconnected from the circuit, and the equivalent resistance is the resistance seen across the terminals when all voltage sources are replaced with short circuits and all current sources are replaced with open circuits.

What is the purpose of finding the Thevenin equivalent?

The purpose of finding the Thevenin equivalent is to simplify a complex circuit and make it easier to analyze. It allows us to determine the behavior of the circuit at a specific set of terminals without having to consider the entire circuit. This is especially useful in circuit analysis and design.

Can the Thevenin equivalent be used for any circuit?

Yes, the Thevenin equivalent can be used for any linear circuit, meaning that the current through and voltage across any element in the circuit is directly proportional to the applied voltage and current, respectively. Non-linear elements such as diodes and transistors cannot be represented by a Thevenin equivalent.

How accurate is the Thevenin equivalent?

The Thevenin equivalent is an approximation of the original circuit, so it is not always 100% accurate. However, it is accurate enough for most practical applications and can greatly simplify circuit analysis. The accuracy of the Thevenin equivalent depends on the complexity of the original circuit and the accuracy of the measurements used to calculate it.

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